Natural Number Multiplication Distributes over Addition

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Theorem

The operation of multiplication is distributive over addition on the set of natural numbers $\N$:

$\forall x, y, z \in \N:$
$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$
$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$


Proof 1

\(\displaystyle \left({x + y}\right) \times z\) \(=\) \(\displaystyle +^z \left({x + y}\right)\) Definition of multiplication
\(\displaystyle \) \(=\) \(\displaystyle \left({+^z x}\right) + \left({+^z y}\right)\) Power of Product of Commuting Elements in Semigroup equals Product of Powers
\(\displaystyle \) \(=\) \(\displaystyle x \times z + y \times z\)

$\Box$

\(\displaystyle z \times \left({x + y}\right)\) \(=\) \(\displaystyle +^{x + y} z\) Definition of multiplication
\(\displaystyle \) \(=\) \(\displaystyle \left({+^x z}\right) + \left({+^y z}\right)\) Index Laws for Semigroup: Sum of Indices
\(\displaystyle \) \(=\) \(\displaystyle z \times x + z \times y\)

$\blacksquare$


Proof 2

We are to show that:

$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$

for all $x, y, z \in \N$.


From the definition of natural number multiplication, we have by definition that:

\(\displaystyle \forall m, n \in \N: \ \ \) \(\displaystyle m \times 0\) \(=\) \(\displaystyle 0\)
\(\displaystyle m \times n^+\) \(=\) \(\displaystyle \left({m \times n}\right) + m\)


Let $x, y \in \N$ be arbitrary.

For all $z \in \N$, let $P \left({z}\right)$ be the proposition:

$\forall x, y \in \N: \left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$


Basis for the Induction

$P \left({0}\right)$ is the case:

\(\displaystyle \left({x + y}\right) \times 0\) \(=\) \(\displaystyle 0\) Definition of Natural Number Multiplication
\(\displaystyle \) \(=\) \(\displaystyle 0 + 0\) Definition of Natural Number Addition
\(\displaystyle \) \(=\) \(\displaystyle x \times 0 + y \times 0\) Definition of Natural Number Multiplication

and so $P \left({0}\right)$ holds.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.


So this is our induction hypothesis:

$\forall x, y \in \N: \left({x + y}\right) \times k = \left({x \times k}\right) + \left({y \times k}\right)$


Then we need to show:

$\forall x, y \in \N: \left({x + y}\right) \times k^+ = \left({x \times k^+}\right) + \left({y \times k^+}\right)$


Induction Step

This is our induction step:


\(\displaystyle \left({x + y}\right) \times k^+\) \(=\) \(\displaystyle \left({x + y}\right) \times k + \left({x + y}\right)\) Definition of Natural Number Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \left({x \times k}\right) + \left({y \times k}\right) + \left({x + y}\right)\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({x \times k}\right) + x}\right) + \left({\left({y \times k}\right) + y}\right)\) Natural Number Addition is Commutative and Associative
\(\displaystyle \) \(=\) \(\displaystyle \left({x \times k^+}\right) + \left({y \times k^+}\right)\) Definition of Natural Number Multiplication

So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction:

$\forall x, y, z \in \N: \left({x + y}\right) \times n = \left({x \times z}\right) + \left({y \times z}\right)$

$\Box$


Next we need to show that:

$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

for all $x, y, z \in \N$.

So:

\(\displaystyle z \times \left({x + y}\right)\) \(=\) \(\displaystyle \left({x + y}\right) \times z\) Natural Number Multiplication is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left({x \times z}\right) + \left({y \times z}\right)\) from above
\(\displaystyle \) \(=\) \(\displaystyle \left({z \times x}\right) + \left({z \times y}\right)\) Natural Number Multiplication is Commutative

Thus we have proved:

$\forall x, y, z \in \N: z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$

$\blacksquare$


Proof 3

In the Axiomatization of 1-Based Natural Numbers, this is rendered:

$\forall x, y, z \in \N_{> 0}:$
$\left({x + y}\right) \times z = \left({x \times z}\right) + \left({y \times z}\right)$
$z \times \left({x + y}\right) = \left({z \times x}\right) + \left({z \times y}\right)$


Using the following axioms:

\((A)\)   $:$     \(\displaystyle \exists_1 1 \in \N_{> 0}:\) \(\displaystyle a \times 1 = a = 1 \times a \)             
\((B)\)   $:$     \(\displaystyle \forall a, b \in \N_{> 0}:\) \(\displaystyle a \times \paren {b + 1} = \paren {a \times b} + a \)             
\((C)\)   $:$     \(\displaystyle \forall a, b \in \N_{> 0}:\) \(\displaystyle a + \paren {b + 1} = \paren {a + b} + 1 \)             
\((D)\)   $:$     \(\displaystyle \forall a \in \N_{> 0}, a \ne 1:\) \(\displaystyle \exists_1 b \in \N_{> 0}: a = b + 1 \)             
\((E)\)   $:$     \(\displaystyle \forall a, b \in \N_{> 0}:\) \(\displaystyle \)Exactly one of these three holds:\( \)             
\(\displaystyle a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \)             
\((F)\)   $:$     \(\displaystyle \forall A \subseteq \N_{> 0}:\) \(\displaystyle \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \)             


Left Distributive Law for Natural Numbers

First we show that:

$n \times \left({x + y}\right) = \left({n \times x}\right) + \left({n \times y}\right)$


Let us cast the proposition in the form:

$\forall a, b, n \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: a \times \paren {b + n} = \paren {a \times b} + \paren {a \times n}$


Basis for the Induction

$\map P 1$ is the case:

\(\displaystyle a \times \paren {b + 1}\) \(=\) \(\displaystyle \paren {a \times b} + a\) Axiom $B$
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \times b} + \paren {a \times 1}\) Axiom $A$

and so $\map P 1$ holds.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\forall a, b \in \N_{> 0}: a \times \paren {b + k} = \paren {a \times b} + \paren {a \times k}$


Then we need to show:

$\forall a, b \in \N_{> 0}: a \times \paren {b + \paren {k + 1} } = \paren {a \times b} + \paren {a \times \paren {k + 1} }$


Induction Step

This is our induction step:

\(\displaystyle a \times \paren {b + \paren {k + 1} }\) \(=\) \(\displaystyle a \times \paren {\paren {b + k} + 1}\) Axiom $C$
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \times \paren {b + k} } + a\) Axiom $B$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {a \times b} + \paren {a \times k} } + a\) Induction hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \times b} + \paren {\paren {a \times k} + a}\) Natural Number Addition is Associative
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \times b} + \paren {a \times \paren {k + 1} }\) Axiom $B$

The result follows by the Principle of Mathematical Induction.

$\Box$


Right Distributive Law for Natural Numbers

Then we show that:

$\left({x + y}\right) \times n = \left({x \times n}\right) + \left({y \times n}\right)$


For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times n = \paren {a \times n} + \paren {b \times n}$


Basis for the Induction

$\map P 1$ is the case:

\(\displaystyle \paren {a + b} \times 1\) \(=\) \(\displaystyle a + b\) Axiom $A$
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \times 1} + \paren {b \times 1}\) Axiom $A$

and so $\map P 1$ holds.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times k = \paren {a \times k} + \paren {b \times k}$


Then we need to show:

$\forall a, b \in \N_{> 0}: \paren {a + b} \times \paren {k + 1} = \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }$


Induction Step

This is our induction step:

\(\displaystyle \paren {a + b} \times \paren {k + 1}\) \(=\) \(\displaystyle \paren {\paren {a + b} \times k} + \paren {a + b}\) Axiom $B$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {a \times k} + \paren {b \times k} } + \paren {a + b}\) Induction hypothesis
\(\displaystyle \) \(=\) \(\displaystyle \paren {\paren {a \times k} + a} + \paren {\paren {b \times k} + b}\) Natural Number Addition is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \times \paren {k + 1} } + \paren {b \times \paren {k + 1} }\) Axiom $B$

The result follows by the Principle of Mathematical Induction.

$\Box$

The result follows.

$\blacksquare$


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